To solve the equation
\[ \sqrt{x^2 + 3x + 2} = \sqrt{x^2 + 3}, \]
we will first square both sides to eliminate the square roots:
\[ x^2 + 3x + 2 = x^2 + 3. \]
Next, we simplify the equation by subtracting \(x^2\) from both sides:
\[ 3x + 2 = 3. \]
Now, we isolate the term involving \(x\) by subtracting 2 from both sides:
\[ 3x = 1. \]
Next, divide both sides by 3:
\[ x = \frac{1}{3}. \]
Now we will verify if this solution satisfies the original equation:
- Calculate \(x^2 + 3x + 2\) for \(x = \frac{1}{3}\):
\[ \left(\frac{1}{3}\right)^2 + 3\left(\frac{1}{3}\right) + 2 = \frac{1}{9} + 1 + 2 = \frac{1}{9} + \frac{9}{9} + \frac{18}{9} = \frac{28}{9}. \]
- Calculate \(x^2 + 3\) for \(x = \frac{1}{3}\):
\[ \left(\frac{1}{3}\right)^2 + 3 = \frac{1}{9} + 3 = \frac{1}{9} + \frac{27}{9} = \frac{28}{9}. \]
Both sides are equal, confirming \(x = \frac{1}{3}\) is indeed a solution.
Thus, the solution is
\[ \boxed{\frac{1}{3}}. \]